Bending Moment and shear force diagram for overhanging beam

Problem 5-3

Determine the values and draw the diagrams for shear force and bending moment due to the imposed loads on overhanging beam shown in figure 5-3(a) and find the position of point of contra-flexure, if any.

Figure 5-3(a)

Solution:

The free-body-diagram is shown in figure 5-3(b). The unknown reaction components of the beam are calculated in Prob 4-2 as  A_x=0,  A_y = -10 kN and B_y = 55 kN;

 Shearing force calculations

 In case of point load acting at a point, we should calculate shear force on both sides (left and right) of the point.

F_A-left = 0;

F_A-right = -10 kN;

F_C = -10 kN;

F_B-left = -10 – 5x4 = - 30 kN;

F_B -right = -10 – 5x4 + 55 = +25 kN;

F_D-left = +25 kN;

F_D-right = +25 – 25 = 0;

 The values of shear force are plotted in SFD in figure 5-3(b)

Bending moment calculations

M_A = 0;

M_C = –10x2 = -20 kNm;

M_B = –10x6 – 5x4x2 = –60 –40 = –100 kNm;

M_D = 0;

The values of bending moment are plotted in BMD.in figure 5-3(b)

Figure 5-3(b)

Maximum bending moment will occur at the point of where shear force changes from negative to positive, which can be easily determined from shear force diagram which tells us that point B is such a point. Therefore the maximum bending moment in this case will occur at B which is 100 kNm.

The bending moment diagram shows that bending moment is hogging throughout the span of the beam and maximum BM is at B (shear force changes sign at B). The bending moment sign does not change, hence there is no point of contra-flexure. The bending moment diagram is straight line from A to C (due to point load), parabolic from C to B (due to udl) and straight line from B to D (due to point load).  It is important to note that the parabola is a increasing parabola (means slope is increasing) because the shear force is increasing from C to B. It is based on the fact that the rate of change of bending moment is equal to shear force.